Problem: The sum of the first $n$ terms in the infinite geometric sequence $\left\{1,\frac{1}{3},\frac{1}{9},\frac{1}{27},\dots \right\}$ is $\frac{121}{81}$. Find $n$.
Solution: This is a geometric sequence with first term $1$ and common ratio $\frac{1}{3}$. Thus the sum of the first $n$ terms is: $$\frac{121}{81}=\frac{1\left[1-\left(\frac{1}{3}\right)^n\right]}{1-\frac{1}{3}}=\frac{3^n-1}{2\cdot 3^{n-1}}.$$Solving we have: \begin{align*}
\frac{3^n-1}{2\cdot 3^{n-1}} &= \frac{121}{81}\\
\frac{3^n-1}{2\cdot 3^n} &= \frac{121}{243}\\
243(3^n - 1) &= 121\cdot2\cdot 3^n\\
243\cdot 3^n - 243 &= 242\cdot 3^n\\
3^n&= 243\\
n&= \boxed{5}.
\end{align*}